# `sum_(n=1)^oo 1/sqrt(n+2)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.

`sum_(n=1)^oo1/sqrt(n+2)`

The integral test is applicable if f is positive , continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.

For the given series `a_n=1/sqrt(n+2)`

Consider `f(x)=1/sqrt(x+2)`

Refer to the attached graph of the function. From the graph we observe that the function is positive and continuous for `x>=1`

Let's determine whether the function is decreasing by finding the derivative `f'(x)`

`f'(x)=(-1/2)(x+2)^(-1/2-1)`

`f'(x)=-1/2(x+2)^(-3/2)`

`f'(x)=-1/(2(x+2)^(3/2))`

`f'(x)<0` which implies that the function is decreasing.

We can apply the integral test,since the function satisfies the conditions for the integral test.

Now let's determine whether the improper integral `int_1^oo1/sqrt(x+2)dx` converges or diverges.

`int_1^oo1/sqrt(x+2)dx=lim_(b->oo)int_1^b1/sqrt(x+2)dx`

Let's first evaluate the indefinite integral `int1/sqrt(x+2)dx`

Apply integral substitution:`u=x+2`

`=>du=dx`

`int1/sqrt(x+2)dx=int1/sqrt(u)du`

Apply the power rule,

`=(u^(-1/2+1)/(-1/2+1))`

`=2u^(1/2)`

`=2sqrt(u)`

Substitute back `u=x+2`

`=2sqrt(x+2)+C`  where C is a constant

`int_1^oo1/sqrt(x+2)=lim_(b->oo)[2sqrt(x+2)]_1^b`

`=lim_(b->oo)[2sqrt(b+2)-2sqrt(1+2)]`

`=2lim_(b->oo)sqrt(b+2)-2sqrt(3)`

`=2(oo)-2sqrt(3)`

`=oo-2sqrt(3)`

`=oo`

Since the integral `int_1^oo1/sqrt(x+2)dx` diverges, we conclude from the integral test that the series also diverges.

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